What is area y=sqrt(x)+1,y=(x^2)/4 ,x=0,x=1 ?

The answer to area y=sqrt(x)+1,y=(x^2)/4 ,x=0,x=1 is 19/12

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area y=sqrt(x)+1,y=(x^2)/4 ,x=0,x=1

area

y = x + 1, y = \frac{x ^{2}}{4}, x = 0, x = 1

\frac{19}{12}

Decimal Notation

1.58333 \dots

Solution steps

Apply the area formula:

\int_{0}^{1} x + 1 - \frac{x ^{2}}{4} d x

Solve

\int_{0}^{1} x + 1 - \frac{x ^{2}}{4} d x : \frac{19}{12}

The area is:

= \frac{19}{12}

y^{^{'}} = \frac{x ^{2} - 3 y ^{2}}{2 x y}

x \to 11 lim (3)

\frac{d}{d x} (3 x - 4 x^{\frac{5}{6}})

\int 3 ln (x^{2} - 1) d x

2 \frac{d y}{d x} + \frac{1}{x} y = \frac{x ^{2}}{y}

What is area y=sqrt(x)+1,y=(x^2)/4 ,x=0,x=1 ?
The answer to area y=sqrt(x)+1,y=(x^2)/4 ,x=0,x=1 is 19/12